No, it would be much harder. It's always easier to expand than to condense.
Expansion of the identities is a very important concept.
cos(2x) = 2cos^2(x) - 1 If it's cos(4x), then that's like cos(2*2x). Everything in the expansion is the same except the x will be 2x now, like 2cos^2(2x) - 1.
All you're doing is repeating the expansion process until you can't do it anymore.
If you have more questions that are confusing you, then I can try to help.
In this line, we're changing only the cos^2(2x). When the squared is written on the "cos" part, that means the entire function is squared. Therefore, cos^2(2x) is the same as (cos(2x))^2.
When something is squared, you multiply it by itself. For example, 4^2 = 4*4, and (n+1)^2 = (n+1)(n+1).
Therefore, we are saying that (cos(2x))^2 = (cos(2x))(cos(2x)) = cos2x * cos2x. The rest of the line remained unchanged.
1) Verify the identity cos^4x - sin^4x = cos2x I know that this is the difference of squares so (cos^2x)^2 - (sin^2x)^2 is this correct for the first step?
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360--cos^2x - 1 = sin^2x--Attempt:cos^2x - 1 - sin^2x = 0cos^2x - 1 - (1 - cos^2x) = 0cos^2x - 1 - 1 + cos^2x = 02cos^2x - 2 = 0(2cos^2x/2)= (-2/2)cos^2x = -1cosx = square root of -1And I can't do anything with this now...what am I doing wrong?--Textbook answers:0, 180, 360--
4. (20 points) Consider the polar curve r = 4cos 6 and r=1+2cos e. 1+2 caso (a) Find ALL intersection points with 0 So< 2x and express them in polar coordi- nates. Show your works (b) Find the area inside the smaller loop of the curve r = 1+2 cose. Show your work. (c) Find the length of r = 4 cose from A to B as 8 increases, where A is the inter- section of the two curves in...
5. By using Fresnel method find f(x)where f(x) = 4 cos(2x) + 3 cos (2x + 2) + 2cos(2-2) 6 Coneider 75 re crte choum in the iouro2 Thie crnte I
5. By using Fresnel method find f(x)where f(x) = 4 cos(2x) + 3 cos (2x + 2) + 2cos(2-2) 6 Coneider 75 re crte choum in the iouro2 Thie crnte I
The identity of:
sin^4x+cos^4x=
(sin^2x+cos^2x)(sin^2x+cos^2x)
sin^2x+cos^2=1
This is the answer I come up with, it is not one of the options available as an answer.
The answers given are
1. -2sin^2xcos^2x
2. 1+2sin^2x-2sin^4x
3. 1+3sin^3x-2sin^2x
4. 1-2sin^2x+2sin^4x
5. 0